Simple Tension Exercise: Acceleration and Final Velocity Calculation

[Kernul]

Homework Statement

A mass $m_1$ is attached to a second mass $m_2$ by an Acme (massless, unstretchable) string. $m_1$ sits on a frictionless table; $m_2$ is hanging over the ends of a table, suspended by the taut string from an Acme (frictionless, massless) pulley. At time $t = 0$ both masses are released.
Find:
a) The acceleration of the two masses.
b) The tension T in the string.
How fast are the two blocks moving when mass $m_2$ has fallen a height $H$ (assuming that $m_1$ hasn’t yet hit the pulley)?

Homework Equations

Newton's Second Law
Tension

The Attempt at a Solution

So, the forces in the first mass are null on the y-axis while it's $T$ on the x-axis. (Putting right as the positive direction of the x-axis and up as the positive direction of the y-axis)
So, we have:
$$F_1 = m_1 a_1 = T$$
$$a_1 = \frac{T}{m_1}$$
And this is the first acceleration.
The second mass has instead two forces acting on it on the y-axis. (nothing on the x-axis since it is just going down) These two forces are $T$ and $P = m_2 g$. Since $P$ goes down, it is a negative force.
So we will have:
$$F_2 = m_2 a_2 = T - m_2 g$$
$$a_2 = \frac{T}{m_2} - g$$
And this is the second acceleration.
Now the problem asks for the tension $T$. Isn't this one just $m_1 a_1$? So I basically ended up answering both at the same time, right?

For the last question, I have to find the final velocity from the initial point until $H$. This should be easy. Finding the final time $t_f$ and then substituting in the motion equation with $\Delta x = H$ we end up with this equation:
$$v_f = \sqrt{2 g H}$$
Is this way of doing correct?

 

[mfb]

You can calculate a₁, it should not appear in the final formula for the tension. In the same way, the tension should not appear in the formulas for the acceleration.
There is one equation that you missed, it is related to the fixed string length.

Your answer for the velocity is wrong, and I don't understand how you got it.

 

[Kernul]

You can calculate a₁, it should not appear in the final formula for the tension. In the same way, the tension should not appear in the formulas for the acceleration.
There is one equation that you missed, it is related to the fixed string length.

You mean the fact that I should take the two masses as one? Like $m = (m_1 + m_2)$ since the string just attaches them together.

Your answer for the velocity is wrong, and I don't understand how you got it.

Ops. I got a sign wrong and thought I did it. Because I wanted to arrive at this equation:
$$v_f^2 - v_0^2 = 2 a \Delta x$$
With $v_0 = 0, \Delta x = H, a = -g$.
But, obviously, I end up with $v_f^2 = - 2 g H$, which is wrong.

 

[mfb]

You mean the fact that I should take the two masses as one? Like $m = (m_1 + m_2)$ since the string just attaches them together.

The total mass can be useful at some point, but that's not the point where you need to introduce it.

Are the two accelerations independent?

Ops. I got a sign wrong and thought I did it. Because I wanted to arrive at this equation:
$$v_f^2 - v_0^2 = 2 a \Delta x$$
With $v_0 = 0, \Delta x = H, a = -g$.
But, obviously, I end up with $v_f^2 = - 2 g H$, which is wrong.

The acceleration is not -g (and not g either). You didn't calculate the acceleration yet.
a and delta x should have the same sign, either both positive or both negative.

 

[Kernul]

The total mass can be useful at some point, but that's not the point where you need to introduce it.

Are the two accelerations independent?

If one of the two accelerates, the other one will too because they are attached by an Acme string. So no, they are not independent.

The acceleration is not -g (and not g either). You didn't calculate the acceleration yet.
a and delta x should have the same sign, either both positive or both negative.

Oh true. And $\Delta x$ would be negative since it's going down.

 

[mfb]

If one of the two accelerates, the other one will too because they are attached by an Acme string. So no, they are not independent.

Express this as formula and it will allow you to calculate everything.

 

[Kernul]

Express this as formula and it will allow you to calculate everything.

This means that the two accelerations are the same? So $a = a_1 = a_2$?

 

[mfb]

Right.

 

[Kernul]

Right.

So $a = \frac{m_2}{m_1 + m_2} g$, $T = \frac{m_1 m_2}{m_1 + m_2} g$, and $v_f = \sqrt{\frac{m_2}{m_1 + m_2} 2 g H}$?

 

[TomHart]

Maybe double-check your result for T.

 

[Kernul]

Maybe double-check your result for T.

Yeah, sorry, I forgot to write the $g$. Now it's fixed.

 

[mfb]

Correct.